Problem: What is the least positive integer value of $x$ such that $(2x)^2 + 2\cdot 37\cdot 2x + 37^2$ is a multiple of 47?
Solution: We note that $(2x)^2 + 2\cdot 37 \cdot 2x + 37^2 = (2x + 37)^2$. In order for this expression to be a multiple of 47, $2x + 37$ must be a multiple of 47. Since we want the least positive value of $x$, we will want $2x + 37 = 47$. It follows that $x = \boxed{5}$.